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Sindrei [870]
3 years ago
8

Polarizing windows, filters, etc. are often used to reduce the amount of light that enters the lens of a camera or into a room o

r a car. A library atrium has an overhead skylight that lets in too much light during the day which heats up the interior of the library far too much. The building engineer installs new double paned polarizing sky lights to reduce the intensity. If sunlight, which is unpolarized, has an average intensity of 1250 W/m^2.
Required:
What angle should the polarizing axis of the second pane of the window make with the polarizing axis of the first pane of the window in order to reduce the intensity of the sunlight to 33% of the original value?
Physics
1 answer:
LekaFEV [45]3 years ago
7 0

Answer:

The answer is "35.6^{\circ}"

Explanation:

The sunlight level of the first panel:

I_1 = \frac{I_o}{2}

When the light of this intensity passes through the second window:

I_2 = I_1 \cos^2 \theta\\\\I_2 = \frac{I_o}{2} \cos^2 \theta

 \frac{I_2}{I_o} = 0.33 (33\%) \\\\

therefore,  

0.33 = \frac{1}{2} \cos^2 \theta\\\\\cos^2 \theta = 0.66\\\\\cos \theta = \sqrt{0.66} = 0.8124\\\\\theta = \cos^{-1}( 0.8124) = 35.6^{\circ}\\\\

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T=2pi square root 1/g solve for g.<br> Explanation would be really helpful.
Natalija [7]

I added individual steps for clarity. Note that g must be positive if the solution is to be real.

T=2\pi \sqrt{\frac{1}{g}}=2\pi g^{-\frac{1}{2}}\\g^{-\frac{1}{2}} = \frac{T}{2\pi}\\(g^{-\frac{1}{2}})^{-2} = (\frac{T}{2\pi})^{-2}\\g = \frac{4\pi^2}{T^2}\,\,\,, g>0}

Let me know if you have any questions.

7 0
3 years ago
Q.Solve the following circuit find total resistance RT. Also find value of voltage across resister RC.
Drupady [299]

Answer:

14.57 ohms

Explanation:

Here in the figure ,Rb & R₄are in series  & also  Rc & R₅ are in series. As they are in series , ( Rb + R₄ ) & (Rc & R₅) are in parallel . So the equivalent resistance in that branch = ( 2 + 18 ) ║ ( 3 + 12 )

                                          = 20 ║ 15

                                          = (20×15) / (20 + 15)

                                          = 8.57 ohms

Also Ra ( 6 ohm ) is in series with that branch ,. So the equivalent resistance of the whole circuit = 8.57 + 6 = 14.57 ohms.

5 0
3 years ago
What happens to the speed of light as it passes into a glass block?
Deffense [45]
<h3>♫ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ♫</h3>

➷ The speed would reduce as glass is denser than air.

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

7 0
3 years ago
Read 2 more answers
At t=0 a grinding wheel has an angular velocity of 25.0 rad/s. It has a constant angular acceleration of 26.0 rad/s2 until a cir
Agata [3.3K]

Answer:

a) The total angle of the grinding wheel is 569.88 radians, b) The grinding wheel stop at t = 12.354 seconds, c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

Explanation:

Since the grinding wheel accelerates and decelerates at constant rate, motion can be represented by the following kinematic equations:

\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

\omega = \omega_{o} + \alpha \cdot t

\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})

Where:

\theta_{o}, \theta - Initial and final angular position, measured in radians.

\omega_{o}, \omega - Initial and final angular speed, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

t - Time, measured in seconds.

Likewise, the grinding wheel experiments two different regimes:

1) The grinding wheel accelerates during 2.40 seconds.

2) The grinding wheel decelerates until rest is reached.

a) The change in angular position during the Acceleration Stage can be obtained of the following expression:

\theta - \theta_{o} = \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

If \omega_{o} = 25\,\frac{rad}{s}, t = 2.40\,s and \alpha = 26\,\frac{rad}{s^{2}}, then:

\theta-\theta_{o} = \left(25\,\frac{rad}{s} \right)\cdot (2.40\,s) + \frac{1}{2}\cdot \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)^{2}

\theta-\theta_{o} = 134.88\,rad

The final angular angular speed can be found by the equation:

\omega = \omega_{o} + \alpha \cdot t

If  \omega_{o} = 25\,\frac{rad}{s}, t = 2.40\,s and \alpha = 26\,\frac{rad}{s^{2}}, then:

\omega = 25\,\frac{rad}{s} + \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)

\omega = 87.4\,\frac{rad}{s}

The total angle that grinding wheel did from t = 0 s and the time it stopped is:

\Delta \theta = 134.88\,rad + 435\,rad

\Delta \theta = 569.88\,rad

The total angle of the grinding wheel is 569.88 radians.

b) Before finding the instant when the grinding wheel stops, it is needed to find the value of angular deceleration, which can be determined from the following kinematic expression:

\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})

The angular acceleration is now cleared:

\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}

Given that \omega_{o} = 87.4\,\frac{rad}{s}, \omega = 0\,\frac{rad}{s} and \theta-\theta_{o} = 435\,rad, the angular deceleration is:

\alpha = \frac{ \left(0\,\frac{rad}{s}\right)^{2}-\left(87.4\,\frac{rad}{s} \right)^{2}}{2\cdot \left(435\,rad\right)}

\alpha = -8.780\,\frac{rad}{s^{2}}

Now, the time interval of the Deceleration Phase is obtained from this formula:

\omega = \omega_{o} + \alpha \cdot t

t = \frac{\omega - \omega_{o}}{\alpha}

If \omega_{o} = 87.4\,\frac{rad}{s}, \omega = 0\,\frac{rad}{s}  and \alpha = -8.780\,\frac{rad}{s^{2}}, the time interval is:

t = \frac{0\,\frac{rad}{s} - 87.4\,\frac{rad}{s} }{-8.780\,\frac{rad}{s^{2}} }

t = 9.954\,s

The total time needed for the grinding wheel before stopping is:

t_{T} = 2.40\,s + 9.954\,s

t_{T} = 12.354\,s

The grinding wheel stop at t = 12.354 seconds.

c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

4 0
3 years ago
the speed of travel of the moon around the earth, using the formula for the speed of a moving object in a circular path
Svetach [21]

Answer: 1018.26 m/s

Explanation:

Approaching the orbit of the Moon around the Earth to a circular orbit (or circular path), we can use the equation of the speed of an object with uniform circular motion:  

V=\sqrt{G\frac{M}{r}}

Where:  

V is the speed of travel of the Moon around the Earth

G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} is the Gravitational Constant

M=5.972(10)^{24} kg is the mass of the Earth

r=384400(10)^{3} m is the distance from the center of the Earth to the center of the Moon

Solving:

V=\sqrt{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{5.972(10)^{24} kg}{384400(10)^{3} m}}

V=1018.26 m/s This is the speed of travel of the Moon around the Earth

5 0
3 years ago
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